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Find two consecutive positive integers such that the square of the larger integer is 19 more than 9 times the smaller integer define your variables and write an equation HURRY HELP PLEASE SHOW STEPS

User LightCC
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1 Answer

4 votes

Answer:

9 and 10

Explanation:

x = 1st positive integer

x+1 = 2nd positive integer

Solve for both x and x+1

(x+1) is the next positive integer after x (Ex. 3 aka 2+1 is next integer after 2).

Larger Integer: x+1 ==> being squared

Smaller Integer: x ==> being multipled by 9

9x + 19 = (x+1)^2 ==> the square of (x+1) is 19 more that 9 multiplied by x.

9x + 19 = (x+1)(x+1) ==> simplify

9x + 19 = x(x+1) + 1(x+1) ==> distribute (x+1) to x and 1.

9x + 19 = (x*x + 1*x) + (1*x + 1*1)

9x + 19 = x^2 + x + x + 1 ==> simplify

9x + 19 = x^2 + 2x + 1

19 = x^2 - 7x + 1 ==> subtract 9x on both sides

Make one side of the equation equal to 0 so you can factor the equation:

0 = x^2 - 7x - 18 ==> subtract both sides by 19

0 = x^2 + 2x - 9x - 18 ==> 2x - 9x = -7x and 2 * (-9) = -18

0 = x*x + 2x - 9x - 18

0 = x(x + 2) - 9(x + 2) ==> factor

0 = (x - 9)(x + 2)

x - 9 = 0 x + 2 = 0

x = 9 x = -2

Out of x = 9 and x = -2, only x = 9 is positive.

Hence, x = 9 and x+1=10.

Hence, the answers are 9 and 10.

User PhilG
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